Question

1.0M KOH, 1.0M K2SO4,1.0M NH4NO3,1.0M KNO3 which of the following aqueous solutions should have the highest boiling point

Answer 1

Answer: An aqueous solution of 1.0 m $K_2SO_4$ should have the highest boiling point.

Explanation:

The expression of elevation in boiling point is given by the equation:

$\Delta T_b=i\times k_b\times m$

where,

$\Delta T_b$ = Elevation in boiling point

i = Van't Hoff factor

$T_b$ = change in boiling point

$k_b$ = boiling point constant

m = molality  of the solution

For the given options:

Option 1:  1.0 m KOH

The equation for the ionization of KOH follows:

$KOH\rightarrow K^++OH^-$

Value of i = 2

So, molal concentration will be = $(1.0)\times 2=2.0m$

Option 2:  1.0 m $K_2SO_4$

The equation for the ionization of $K_SO_4$ follows:

$K_2SO_4\rightarrow 2K^++SO_4^{2-}$

Value of i = 3

So, molal concentration will be = $(1.0)\times 3=3.0m$

Option 3:  1.0 m $NH_4NO_3$

The equation for the ionization of $NH_4NO_3$ follows:

$NH_4NO_3\rightarrow NH_4^++NO_3^-$

Value of i = 2

So, molal concentration will be = $(1.0)\times 2=2.0m$

Option 4:  1.0 m $KNO_3$

The equation for the ionization of $KNO_3$ follows:

$KNO_3\rightarrow K^++NO_3^-$

Value of i = 2

So, molal concentration will be = $(1.0)\times 2=2.0m$  

As, the molal concentration of $K_2SO_4$ is the highest, so its boiling point will be the highest.

Hence, an aqueous solution of 1.0 m $K_2SO_4$ should have the highest boiling point.